1st Order RL Transient Problem

Discussion in 'Homework Help' started by bluetooth tamer, Mar 26, 2015.

  1. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
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    Hi,

    I spent some time looking into how to find tau (time constant) of this circuit shown in the professor's handwritten problem and couldn't fully grasp whether he made a mistake or I had. Could some be so kind to help me out with the problem.

    Here's the problem copied and worked out by me.
    IMAG0168.jpg


    Here's my professors work. The problem is the one underneath the line.
    Capture.PNG
    My concern is that the time constants don't match up. I have tau = 4/9 and he has it = 2/9. Which is right?
    My guess is that I didn't make any mistakes and he copied the time constant form the problem above the line by accident into the problem below the line.
     
  2. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
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    Whoa, sorry for the large image. I didn't think the entire 4 mp would scale almost all the way.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes the mistake is presumably yours. You have the switch closing at t=0 sec, whereas it is stated in your prof's solution that the switch has been closed for sufficient time before t=0 to allow the current to reach steady state. The switch is then opened at t=0 sec.
     
  4. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
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    I think that the intention for the second part of the problem was to change the condition for the switch. The instructions say that if the switch has already been open while dc steady state then find i(t) when t >0 where at t =0 the switch closes. Does this argument address what you meant?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    You both need to be slapped for sloppy treatment of your units -- your instructor needs to be slapped much harder just because they are the instructor.

    To find the time constant you want to find the resistance seen by the inductor for the circuit as it is configured at that moment.

    When the switch is open, the inductor sees (3Ω+6Ω)=9Ω of resistance and so the time constant is tau=L/R=2H/9Ω=(2/9)s.

    When the switch is closed, the inductor sees (3Ω+(6Ω||2Ω))=(3Ω+(12/8)Ω)=(9/2)Ω of resistance and so the time constant is tau=L/R=2H/((9/2)Ω)=(4/9)s.
     
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  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think we should have a new video made for the video forum.
    I'd like to suggest the title - "The Wrath of WBahn" - with the none too subtle Star Trek reference.
    You would, of course, take the lead role.

    Apologies to @bluetooth tamer about this hijack. And yes, I missed your point about the second part of the problem. Unfortunately - both you and the prof got that solution wrong. A worrying outcome.
     
    Last edited: Mar 26, 2015
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  7. WBahn

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    Mar 31, 2012
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    I'm envisioning a parody of a couple of scenes from the movie and I think it could be hilarious!
     
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  8. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
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    Lol guys what's wrong in my solution?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Nothing, apart from the units issue.
    I'm too in awe of WBahn to cross him on that score.
    At least you spotted the prof's error & to your credit didn't accept his solution on blind faith.
     
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